A sequence \(\{u_n\}\) is arithmetic if \[ u_{n+1} - u_n = d \quad \text{for all } n, \] where \(d\) is a constant called the common difference.

Notation:

• First term: \(u_1 = a\).
• Common difference: \(d\).
• General term (nth term): \[ u_n = a + (n - 1)d, \quad n = 1,2,3,\dots \]

Example 1

The sequence \(4,\, 7,\, 10,\, 13,\, \dots\) is arithmetic.

• \(a = 4\)
• \(d = 7 - 4 = 3\)
• \(\displaystyle u_n = 4 + (n - 1)\cdot 3 = 3n + 1\).

For example, \(u_5 = 3 \times 5 + 1 = 16\).

Let \(S_n\) be the sum of the first \(n\) terms of an arithmetic sequence: \[ S_n = u_1 + u_2 + \dots + u_n. \]

Writing the sum forwards and backwards:

\[ \begin{aligned} S_n &= a + (a + d) + (a + 2d) + \dots + [a + (n-1)d], \\ S_n &= [a + (n-1)d] + [a + (n-2)d] + \dots + a. \end{aligned} \]

Adding term by term gives \(n\) pairs, each equal to \(2a + (n-1)d\): \[ 2S_n = n\bigl(2a + (n - 1)d\bigr). \] So \[ S_n = \frac{n}{2}\bigl(2a + (n - 1)d\bigr). \]

Example 2

For the sequence \(4, 7, 10, 13, \dots\) from Example 1, find:

• (i) the 20th term,
• (ii) the sum of the first 20 terms.

Solution.

(i) \(u_n = 3n + 1\), so \[ u_{20} = 3(20) + 1 = 61. \]

(ii) \(a = 4\), \(d = 3\), \(n = 20\). \[ \begin{aligned} S_{20} &= \frac{20}{2}\bigl(2\cdot 4 + (20 - 1)\cdot 3\bigr) \\ &= 10(8 + 57) \\ &= 10 \times 65 \\ &= 650. \end{aligned} \]

So the 20th term is \(\boxed{61}\) and the sum of the first 20 terms is \(\boxed{650}\).

Often you are given two terms of an arithmetic sequence, for example \(u_p\) and \(u_q\). Using \[ u_n = a + (n - 1)d \] you can form two equations and solve for \(a\) and \(d\).

Example 3

In an arithmetic sequence, the 4th term is 11 and the 9th term is 26.

• (i) Find \(a\) and \(d\).
• (ii) Find the sum of the first 15 terms.

Solution.

(i) Using \(u_n = a + (n - 1)d\): \[ \begin{aligned} u_4 &= a + 3d = 11, \\ u_9 &= a + 8d = 26. \end{aligned} \] Subtract the first from the second: \[ (a + 8d) - (a + 3d) = 26 - 11 \quad\Rightarrow\quad 5d = 15 \Rightarrow d = 3. \] Substitute into \(a + 3d = 11\): \[ a + 3\cdot 3 = 11 \Rightarrow a = 2. \] So \(a = 2\), \(d = 3\).

(ii) Sum of first 15 terms: \[ \begin{aligned} S_{15} &= \frac{15}{2}\bigl(2a + (15 - 1)d\bigr) \\ &= \frac{15}{2}\bigl(2\cdot 2 + 14\cdot 3\bigr) \\ &= \frac{15}{2}(4 + 42) \\ &= \frac{15}{2} \cdot 46 \\ &= 15 \cdot 23 \\ &= 345. \end{aligned} \]

So \(S_{15} = \boxed{345}\).

Sometimes you are given sums rather than terms, for example \(S_n\) and \(S_m\). You can use \[ S_n = \frac{n}{2}\bigl(2a + (n - 1)d\bigr) \] to form equations in \(a\) and \(d\).

Example 4

For an arithmetic sequence, the sum of the first 10 terms is 85 and the sum of the first 20 terms is 260. Find the first term and the common difference.

Solution.

Using \[ S_n = \frac{n}{2}\bigl(2a + (n - 1)d\bigr), \] we have \[ \begin{aligned} S_{10} &= \frac{10}{2}\bigl(2a + 9d\bigr) = 5(2a + 9d) = 85, \\ S_{20} &= \frac{20}{2}\bigl(2a + 19d\bigr) = 10(2a + 19d) = 260. \end{aligned} \]

So \[ \begin{aligned} 2a + 9d &= 17, \quad\text{(divide the first equation by 5)} \\ 2a + 19d &= 26. \quad\text{(divide the second equation by 10)} \end{aligned} \] Subtract: \[ (2a + 19d) - (2a + 9d) = 26 - 17 \Rightarrow 10d = 9 \Rightarrow d = \frac{9}{10}. \] Substitute into \(2a + 9d = 17\): \[ 2a + 9\cdot\frac{9}{10} = 17 \Rightarrow 2a + \frac{81}{10} = 17 \Rightarrow 2a = 17 - \frac{81}{10} = \frac{170 - 81}{10} = \frac{89}{10}, \] so \[ a = \frac{89}{20}. \]

Therefore \(a = \boxed{\dfrac{89}{20}}\) and \(d = \boxed{\dfrac{9}{10}}\).

Example 5

Seats in a small theatre are arranged in rows. The first row has 12 seats, the second row has 14 seats, the third row has 16 seats and so on, forming an arithmetic sequence.

1. Find an expression for the number of seats in the \(n\)th row.
2. Find the number of seats in the 20th row.
3. Find the total number of seats in the first 20 rows.

Solution.

First term \(a = 12\), common difference \(d = 14 - 12 = 2\).

(i) Number of seats in the \(n\)th row: \[ u_n = a + (n - 1)d = 12 + 2(n - 1) = 2n + 10. \]

(ii) 20th row: \[ u_{20} = 2(20) + 10 = 50. \]

(iii) Total number of seats in the first 20 rows: \[ \begin{aligned} S_{20} &= \frac{20}{2}(u_1 + u_{20}) \\ &= 10(12 + 50) \\ &= 10 \times 62 \\ &= 620. \end{aligned} \]

So there are \(\boxed{620}\) seats in the first 20 rows.